Charge densities and delta functions

1. Sep 16, 2016

spaghetti3451

Note from mentor: this thread was originally posted in a non-homework forum, therefore it lacks the homework template.

I was wondering if the electric charge density $\rho({\bf{r}})$ of a point charge $q$ at position ${\bf{r}}_{0}$ is given by $\rho({\bf{r}})=q\delta^{(3)}({\bf{r}}-{\bf{r}}_{0})$.

If so, is this because $\int d^{3}{\bf{r}}\ \rho({\bf{r}}) = \int d^{3}{\bf{r}}\ q\delta^{(3)}({\bf{r}}-{\bf{r}}_{0}) = q$?

2. Sep 16, 2016

vanhees71

If it's a point charge at rest, yes. Note that $\delta^{(3)}(\vec{r}-\vec{r}_0)$ is NOT a function BUT a distribution (generalized function)!

3. Sep 16, 2016

spaghetti3451

Let's say that now we have a point charge $-q$ at the origin and a point charge $+q$ at position $\bf{r}_{0}$. Is the charge density of the dipole given by $\rho({\bf{r}})=-q\delta^{(3)}({\bf{r}})+q\delta^{(3)}({\bf{r}-\bf{r}_{0}})$?

4. Sep 16, 2016

vanhees71

Yes, but usually a dipole is defined as the approximation that $r_0$ is very small in the sense of the multipole expansion.

5. Sep 16, 2016

spaghetti3451

Thanks.

I'm also trying to find the charge density of a uniform infinitesimally thin spherical shell of radius $R$ and total charge $Q$ centred at the origin.

This is my attempt:

$\int\ \rho({\bf{r}})\ dV = Q$

$\lim\limits_{\epsilon\to 0}\ \int\limits_{R-\epsilon}^{R+\epsilon}\ \rho({\bf{r}})\ (\text{sin}\theta\ d\theta\ d\phi)\ (r^{2}dr) = Q$

$\lim\limits_{\epsilon\to 0}\ \int\limits_{R-\epsilon}^{R+\epsilon}\ \rho({\bf{r}})\ (r^{2}dr) = \frac{Q}{4\pi}$

Any hints on what to do next?

6. Sep 21, 2016

rude man

How about assuming ρ(r) = f(r) δ(r - r0) in your last integral
because you know ρ will have to have an infinity at r0 to cancel out the dr in Q = ∫ρ(r) dV = ∫ρ(r) 4π r2 dr
then putting that in your last integral and solving for f(r0) and stating f(r) = f(r0) with r0 = r.

PS I used r0 in lieu of your R.

Last edited: Sep 21, 2016
7. Nov 22, 2016

garylau

Sorry

What is the reason for ρ should have an infinity at r?????

8. Nov 22, 2016

TSny

ρ is infinite at r0.

It's because the shell of charge is modeled as having zero thickness. Since the shell contains a finite amount of charge, the density of charge must be infinite.

It's somewhat similar to a point charge. If a finite amount of charge occupies a single point, then the density of charge must be infinite.

9. Nov 22, 2016

garylau

Do you mean the volume is zero So the Q/volume=infinity?

10. Nov 22, 2016

TSny

Yes, the volume is zero. So the volume charge density ρ is infinite. The surface charge density is finite.

11. Nov 22, 2016

garylau

i See thank you

12. Nov 22, 2016

TSny

You are welcome. I'm glad I could help.

13. Nov 22, 2016

garylau

there are one more question

How to use ρ at infinity to cancel out the dr in Q??

can you explain more

thank

14. Nov 22, 2016

TSny

I'm not sure I understand the question. The idea is that ρ(r) is zero everywhere except where r = R. (I'm switching the notation from r0 to R for the radius of the spherical shell in order to conform to post # 5. Sorry for the notational confusion.) At r = R, ρ is infinite.

So, assume that ρ(r) is proportional to δ(r - R): ρ(r) = Cδ(r - R).

You just need to determine C. Do this by requiring ∫ρ(r)dV = Q for integration over all space. Express the integral in spherical coordinates and work out the value of C.

15. Nov 22, 2016

garylau

it look likes the C should be the surface charge density

16. Nov 22, 2016

TSny

Yes, that's right.

17. Dec 2, 2016

EricGetta

Usually a dipole is formed as a limit as seperation d->0 but with charge q/d to compensate.